1962 AHSME Problems/Problem 20

Problem

The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be:

$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 36$

Solution

If the angles are in an arithmetic progression, they can be expressed as $a$ , $a+n$ , $a+2n$ , $a+3n$ , and $a+4n$ for some real numbers $a$ and $n$ . Now we know that the sum of the degree measures of the angles of a pentagon is $180(5-2)=540$ . Adding our expressions for the five angles together, we get $5a+10n=540$ . We now divide by 5 to get $a+2n=108$ . It so happens that $a+2n$ is one of the angles we defined earlier, so that angle must have a measure of $\boxed{108\textbf{ (A)}}$ . (In fact, for any arithmetic progression with an odd number of terms, the middle term is equal to the average of all the terms.)

Solution 2

If we write the five terms as $a$ , $a - n$ , $a - 2n$ , $a + n$ and $a + 2n$ , we can see that adding them up, we get $5a = 540$ through this, we can see that $a = 108$ , $\fbox{\textbf{(A)}}$

1962 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1962 AHSME Problems/Problem 20

Contents

Problem

Solution

Solution 2

See Also